As seen earlier, work is related to force and the displacement over which it acts. Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive x-direction as shown in Fig. 1.
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| Fig 1: An object undergoes a displacement d under the influence of the force F. |
The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus
W = (F cos θ)d = F.d
We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, the force you exert on the wall does no work. Yet your muscles are alternatively contracting and relaxing and internal energy is being used up and you do get tired. Thus, the meaning of work in physics is different from its usage in everyday language.
No work is done if:
(i) the displacement is zero as seen in the example above. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time.
(ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement.
(iii) the force and displacement are mutually perpendicular. This is so since, for θ = π/2 rad (= 90$^o$), cos (π/2) = 0. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement. If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and θ = π/2.
Work can be both positive and negative. If θ is between 0o and 90$^o$, cos θ in Eq. (6.4) is positive. If θ is between 90o and 180$^o$ , cos θ is negative. In many examples the frictional force opposes displacement and θ = 180$^o$ . Then the work done by friction is negative (cos 180$^o$ = –1).
From Eq. (6.4) it is clear that work and energy have the same dimensions, [M$L^2T^{–2}$]. The SI unit of these is joule (J), named after the famous British physicist James Prescott Joule (1811-1869). Since work and energy are so widely used as physical concepts, alternative units abound and some of these are listed in Table 1.
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| Table: Alternative Units of Work/Energy in J |
Example 1
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion.
(a) How much work does the road do on the cycle ?
(b) How much work does the cycle do on the road?
Answer
Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
(a) The stopping force and the displacement make an angle of 180$^o$ (π rad) with each other. Thus, work done by the road,
Wr = Fd cosθ
= 200 × 10 × cos π = – 2000 J
It is this negative work that brings the cycle to a halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement.
Thus, work done by cycle on the road is zero.
The lesson of Example 1 is that though the force on a body A exerted by the body B is always equal and opposite to that on B by A (Newton’s Third Law); the work done on A by B is not necessarily equal and opposite to the work done on B by A.
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